What is the total energy for incompressible fluid?

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For compressible fluid, the total energy of per unit mass of fluid is

H=CpT+1/2*V^2

From this equation, we can write other variations. There should be no any question for that.

I write in a manuscript:

For incompressible fluid, the total energy of per unit volume of fluid is

p+1/2*rho*V^2+rho*g*z ...................... (1)

This is from the text book of fluid mechanics.

Bernoulli equation means that the quantity expressed by Equation (1) is constant along a streamline in inviscid fluid.

We can refer to:

http://www.centennialofflight.gov/es...vation/TH8.htm

However, The Reviewer of the Journal writes: "This is a Bernoulli construct but is not energy as far as I can tell."

Another Reviewer commented the same: "This is the Bernoulli head, but not the energy."

Can anybody comment for this?

What is the total energy for incompressible fluid?

> What is the total energy for incompressible fluid?

Without more context it is not possible to answer this question.

The problem is that one would need to know what you want "total energy" to represent for your incompressible fluid. The assumption of incompressibility makes the fluid's thermodynamic state inconsistent. Nonetheless one can often sort out mechanical quantities without too much problem.

I mean the total mechanical energy of per unit volume fluid in incompressible flow.

> I mean the total mechanical energy of per unit volume fluid in incompressible flow.

This is a different quantity. A widely held book which derives and compares transport equations for this quantity and energy is "Transport Phenomena" by Bird, Steward and Lightfoot. Depending on context, you might still need to be careful about quantities with a thermodynamic meaning for your incompressible fluid but so long as whatever you are trying to argue is mechanical you probably will be able to make a case your reviewers are happy with. Again, without context it is difficult to know.

The standard definition of total energy is: internal energy + kinetic energy . your H=CpT+1/2*V^2 is actually total enthalpy/mass = CvT+ 1/2*V^2 + P/rho, it is not total energy. For steady inviscid flow, the total enthalpy/volume is constant along a streamline. This is equivalent to Bernoulli equation: p+1/2*rho*V^2+rho*g*z for incompressible flow with gravity into account.

I have given the context in my first post. Is it not appropriate?

It is:

For compressible flow, the total energy is

H=CpT+1/2*rho*V^2

My purpose is to express the counterpart for this term in incompressible flow.

Thanks.

Harry

hi....

the equation u have written comes from the integration of elemental momentum eqn dp/rho+v dv+g dz=0 with the assumption of incompressibility.for a compressible fluid pressure is not just a mechanical parameter but is a state varible.So integrate the above eqn with the help of eqn of state

regards

kasyap

Your expression is total mechanical energy derived from the momentum equation for incompressible flow.

Did the total energy you mean include the thermal energy used in the energy equation?

However, there is a link between the momentum equation and the energy equation in deriving Bernoulli equation.

In my opinion, the total energy (include thermal energy) may be extracted to clerify the physical meanings of all terms in the governing equation. Only some applications require the formulation based on the total energy.

I may be wrong. Please discuss with other person.

> I have given the context in my first post. Is it not appropriate?

It does not explain the point you are trying to make from which one might be able to deduce what you are trying to equate in both compressible and incompressible flow (assuming this is what you are trying to do).

The problem with equating themodynamic properties in compressible flow with their "equivalent" in incompressible flow is that the general thermodynamic relations do not hold for the latter. Some reduced relations might depending on exactly what you are doing.

What can often work are arguments based on mechanical quantities like work so long as you do not push the meaning of pressure too hard.

I guess what the referee means is you should choose 1/2*rho*V^2 as energy. Because the flow is incompressible, pressure shouldn't appear in the energy. And also, the internal energy doesn't have a feedback on the momentum equation as it does in the comperssible case, thus internal energy is always not considered in the incompressible flow.

Just my opinion.

Pete

Thanks. but

> For steady inviscid flow, the total enthalpy/volume is constant along a streamline.

For steady VISCOUS flow, the total enthalpy/volume is constant along a streamline too.

> For steady VISCOUS flow, the total enthalpy/volume is constant along a streamline too.

This is a commonly held view but it is wrong. The viscous term in energy transport equations has two components: the first is the irreversible transfer to internal energy and the second is the reversible work done on the fluid element. If there is a gradient in the latter in the direction between streamlines there will be a transfer of energy between the streamlines.

Boundary layers provide good examples because the viscous forces are high and gradients in the shear stress will generally act to transfer energy between adjacent streamlines. Curving the boundary layer and/or working it against and adverse pressure gradient will induce nice gradients in the shear stress and make the effects apparent.

What you write is total enthalpy not total energy. Total energy is

E=CvT+1/2*rho*V^2

If you want to express

H=CpT+1/2*rho*V^2

in incompressible flow use thermodynamic relation

TdS=dH-dp/rho

and set rho=const